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3 years ago

What is the mass of a football in free fall with a total force due to gravity of 782N?

Physics

2 answers:

Alla [95]3 years ago

8 0

P = 782 N

g = 9.81 m/s^2

So:

m = P/g = 782N/9.81 m/s^2 = 79,7 Kg

Mama L [17]3 years ago

4 0

Explanation:

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A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

Umnica [9.8K]

Answer:

A = 6.9 cm

Explanation:

Simple Harmonic Motion

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

$\displaystyle w=\sqrt{\frac{k}{m}}$

The equation for the motion of the object is written as a sinusoid:

$\displaystyle X=A\ cos\ w\ t$

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

$\displaystyle X'=V=-A\ w\ sin\ w\ t$

And the maximum speed is

$\displaystyle V_{max}= A\ w$

Solving for the amplitude

$\displaystyle A= \frac{V_{max}}{w}$

Computing w

$\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\ rad/ s$

Calculating A

$\displaystyle A=\frac{1.7}{24.5}=0.069\ m$

$\displaystyle \boxed{A=6.9\ cm}$

7 0

3 years ago

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. Rob's destination is a star sys

olchik [2.2K]

Answer:

A) 15.0 years

Explanation:

Due to the distance to the star system is in light-year units, we can compute the time by using:

$t=\frac{d}{v}=\frac{14.4 l-y}{0.960}=15l-y$

then, Rob will take to complete the trip about 15 light-years.

hope this helps!!

7 0

3 years ago

The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.

lesya [120]

Answer:

(a) m = 1.6 x 10²¹ kg

(b) K.E = 2.536 x 10¹¹ J

Explanation:

(a)

First we find the volume of the continent:

V = L*W*H

where,

V = Volume of Slab = ?

L = Length of Slab = 4450 km = 4.45 x 10⁶ m

W = Width of Slab = 4450 km = 4.45 x 10⁶ m

H = Height of Slab = 31 km = 3.1 x 10⁴ m

Therefore,

V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)

V = 6.138 x 10¹⁷ m³

Now, we find the mass:

m = density*V

m = (2620 kg/m³)(6.138 x 10¹⁷ m³)

m = 1.6 x 10²¹ kg

(b)

The kinetic energy will be:

K.E = (1/2)mv²

where,

v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)

v = 3.17 x 10⁻¹⁰ m/s

Therefore,

K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²

K.E = 2.536 x 10¹¹ J

(c)

For the same kinetic energy but mass = 77 kg:

K.E = (1/2)mv²

2.536 x 10¹¹ J = (1/2)(77 kg)v²

v = √(2)(2.536 x 10¹¹ J)

v = 7.12 x 10⁵ m/s

7 0

3 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

$v^2=u^2+2gh$

$h =\dfrac{v^2}{2g}$ ....(I)

For second half,

We need to calculate the time

Using equation of motion

$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

$t_{2}=\sqrt{\dfrac{2h}{g}}$

Put the value of h from equation (I)

$t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}$

$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

The initial velocity is

$v=\dfrac{10\times10}{2}$

$v=50\ m/s$

Hence, The initial velocity is 50 m/s.

3 0

3 years ago